Mariusz Iwaniuk

1491 Reputation

14 Badges

8 years, 281 days

Social Networks and Content at Maplesoft.com

MaplePrimes Activity


These are answers submitted by Mariusz Iwaniuk

w := proc (x, y) options operator, arrow; piecewise(y <= .5, -2*tanh(y-.25), .5 < y, 2*tanh(.75-y)) end proc;
Matrix([seq([seq(w(x, y), x = 0 .. 10)], y = 0 .. 10)]);

#or;

Matrix([seq([seq(w(x, y), y = 0 .. 10)], x = 0 .. 10)]);

 

 

You don't gives values of constans,so I assuming.

restart;
Digits := 20;
Theta := (1/3)*Pi; Upsilon := 1/10;#assume!
eq := ((D@@2)(u))(r) = (-(D(u))(r)^2*u(r)+((Upsilon-1)*(1/2))*(1-u(r)^2-(D(u))(r)^2)*((D(u))(r)*cot(Theta)+2*u(r)))/((D(u))(r)^2-((Upsilon-1)*(1/2))*(1-u(r)^2-(D(u))(r)^2));
sol := dsolve({eq, u(13.75) = .7787, (D(u))(13.75) = .344037}, numeric, abserr = 1.*10^(-16));
plots:-odeplot(sol, [[r, u(r)], [r, (D(u))(r)]], r = 0 .. 13.75, legend = [typeset("Curve: ", u(r)), typeset("Curve: ", (D(u))(r))]);

 

 

It's possible Bug in Ode Analyzer Assistant.

You must write equation like so:

y[1](x) -> y1(x)
y[2](x) -> y2(x)
diff(y1(x), x$2)+3*(diff(y1(x), x$1))+2*y1(x)+2*y2(x) = 3,

diff(y2(x), x$2)+diff(y2(x), x$1)+2*y2(x)+2*y1(x) = 8;

Regards Mariusz.

 

maybe like so:

expand(eval(pdsolve({diff(u(x, z, t), z)+C = 0, u(x, a, t) = 0}), a = h(x, t)));

#u(x, z, t) = C*h(x, t)-C*z

 

 

Maple has this function Totient(n) built-in.This command was introduced in Maple 2016. Using showstat function we can preview the code.

with(NumberTheory):
showstat(Totient);
showstat(PrimeFactors);
NumberTheory:-Totient := proc(n::{posint, And(algebraic,Not({boolean, `in`, complexcons, extended_numeric}))}, $)
local prime_factors, p;
   1   if not type(n,'posint') then
   2       return ('procname')(n)
       elif isprime(n) then
   3       return n-1
       else
   4       prime_factors := NumberTheory:-PrimeFactors(n);
   5       return n*mul(p-1,`in`(p,prime_factors))/convert(prime_factors,'`*`')
       end if
end proc

NumberTheory:-PrimeFactors := proc(n::{integer, And(algebraic,Not({boolean, `in`, complexcons, extended_numeric}))}, $)
local i;
   1   if n = 0 then
   2       error "cannot represent all prime factors of %1", n
       elif type(n,{'negint', 'posint'}) then
   3       return {seq(i[1],`in`(i,ifactors(n)[2]))}
       else
   4       return ('procname')(n)
       end if
end proc

Another code you can find on this webpage: http://oeis.org/A000010 see:MAPLE -> # version 2.

Have fun !

If yours integral is:

int(sqrt(ln(x)/x^2-1/x), x)

then Maple can't find it. Possible it has No closed form.

See: https://en.wikipedia.org/wiki/Nonelementary_integral.

Mathematica, Rubi,Axiom,SymPy, Maxima  cannot do it, either.

Second one:

int((ln(x)^(a-1)/x^2-1/x)^(1/2),x)

the same case.

EDITED:

If You want approximate integral by series see worksheet.

Approximate_indefine_integral_by_Series.mw

convert(exp(x), Sum, dummy = n)

or:

convert(exp(x), FormalPowerSeries)

For Simple function (2 methods):

you can use inverse ztransform:

Sum(x^n*invztrans(eval(exp(x), x = 1/x), x, n), n = 0 .. infinity)#only works if invztrans can find transfrom.

or n-th derivative:

Sum(x^n*(eval(diff(exp(x), x$n), x = 0))/factorial(n), n = 0 .. infinity)#only works if n-th derivative can find.

 

Only another way to calculate Julian date.

Formula from:http://aa.usno.navy.mil/faq/docs/JD_Formula.php

Compute the JD corresponding to 2018 January 11, 18h30m30s UT.

Substituting Y = 2018, M = 1, D = 11,h = 18, m = 30, s = 30;

restart:
JD := proc (Y, M, D, h, m, s) local x; 
x := evalf(367*Y-trunc((7/4)*Y+(7/4)*trunc((1/12)*M+3/4))+trunc((275/9)*M)+D+1721013.5-
(1/2)*signum(100*Y+M-190002.5)+1/2+(1/24)*h+(1/1440)*m+(1/86400)*s); end proc:

JD(2018, 1, 11, 18, 30, 30);

# 2.458130271*10^6

 

 

It may not possible to find anti-derivatives(closed form).

https://math.stackexchange.com/questions/1469123/integral-of-ex2

https://math.stackexchange.com/questions/168860/functions-cannot-be-integrated-as-simple-functions?rq=1

Mathematica gives No solution.

An aproximation only can be done e.g by series:


 

p := sqrt(ln(t)*(1/ln(t)^2))/(t*(1-ln(t)^2/t^2)^(1/4))

(1/ln(t))^(1/2)/(t*(1-ln(t)^2/t^2)^(1/4))

(1)

with(IntegrationTools):

Int(p, t)

Int((1/ln(t))^(1/2)/(t*(1-ln(t)^2/t^2)^(1/4)), t)

(2)

Change(Int((1/ln(t))^(1/2)/(t*(1-ln(t)^2/t^2)^(1/4)), t), t = exp(x))

Int(-(1/ln(exp(x)))^(1/2)*(exp(x))^2*(-(ln(exp(x))^2-(exp(x))^2)/(exp(x))^2)^(3/4)/(ln(exp(x))^2-(exp(x))^2), x)

(3)

`assuming`([simplify(Int(-(1/ln(exp(x)))^(1/2)*(exp(x))^2*(-(ln(exp(x))^2-(exp(x))^2)/(exp(x))^2)^(3/4)/(ln(exp(x))^2-(exp(x))^2), x))], [x > 0])

Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x)

(4)

GetIntegrand(Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x))

exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2))

(5)

series(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x = 0, 5)

1/x^(1/2)+(1/4)*x^(3/2)-(1/2)*x^(5/2)+(21/32)*x^(7/2)+O(x^(9/2))

(6)

Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x) = int(1/x^(1/2)+(1/4)*x^(3/2)-(1/2)*x^(5/2)+(21/32)*x^(7/2)+O(x^(9/2)), x)

Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x) = (1/1680)*x^(1/2)*(245*x^4-240*x^3+168*x^2+3360)+O(x^(11/2))

(7)

``


 

Download Only_aproximation.mw

 

Use Maple built function intsolve with method Neumann or with this  paper https://arxiv.org/ftp/arxiv/papers/1309/1309.6311.pdf see another attached file (Fredholm_integral_2ver.)


 

restart

alpha := 1; A := 1

f := proc (k) options operator, arrow; 1-2*k^2/(sqrt(alpha^2+k^2)*(sqrt(alpha^2+k^2)+k)) end proc

proc (k) options operator, arrow; 1-2*k^2/(sqrt(alpha^2+k^2)*(sqrt(alpha^2+k^2)+k)) end proc

(1)

K := proc (x, t) options operator, arrow; Int(2*f(v)*cos(x*v)*cos(t*v)/Pi, v = 0 .. A) end proc

proc (x, t) options operator, arrow; Int(2*f(v)*cos(x*v)*cos(t*v)/Pi, v = 0 .. A) end proc

(2)

EQ := h(x) = 4/(Pi*alpha^2)-(Int(h(t)*K(x, t), t = 0 .. 1))

h(x) = 4/Pi-(Int(h(t)*(Int(2*(1-2*v^2/((v^2+1)^(1/2)*((v^2+1)^(1/2)+v)))*cos(x*v)*cos(t*v)/Pi, v = 0 .. 1)), t = 0 .. 1))

(3)

HT := intsolve(EQ, h(x), method = Neumann, order = 1)

h(x) = 4*(Int(Int(-2*(-v^2+1+(v^2+1)^(1/2)*v)*cos(x*v)*cos(_k1*v)/((v^2+1)^(1/2)*((v^2+1)^(1/2)+v)), v = 0 .. 1), _k1 = 0 .. 1)+Pi)/Pi^2

(4)

help("intsolve # See Maple Help for more information.")

(1/4)*D/(Pi*mu*U[0]) = int(rhs(HT), x = 0 .. 1, numeric)

(1/4)*D/(Pi*mu*U[0]) = .7334429187

(5)

NULL


 

Download _Fredholm_integral.mw

_Fredholm_integral_2ver.mw

Only for k=1,4,5,6,7 is real solution.


 

restart

sol := allvalues(solve(10*c*x^7-6*x^3+6 = 0, x))

RootOf(5*_Z^7*c-3*_Z^3+3, index = 1), RootOf(5*_Z^7*c-3*_Z^3+3, index = 2), RootOf(5*_Z^7*c-3*_Z^3+3, index = 3), RootOf(5*_Z^7*c-3*_Z^3+3, index = 4), RootOf(5*_Z^7*c-3*_Z^3+3, index = 5), RootOf(5*_Z^7*c-3*_Z^3+3, index = 6), RootOf(5*_Z^7*c-3*_Z^3+3, index = 7)

(1)

Matrix([seq([series(sol[k], c = 0, 5)], k = 1 .. 7)])

Matrix([seq([convert(series(sol[k], c = 0, 5), polynom)], k = 1 .. 7)])

Matrix([[1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [-1/2+((1/2)*I)*3^(1/2)-(5/9)*(-1+I*3^(1/2))*c/(I*3^(1/2)+1)+(50/27)*c^2-(38000/2187)*c^3/(I*3^(1/2)+1)+(1868750/19683)*c^4/(-1+I*3^(1/2))], [-1/2-((1/2)*I)*3^(1/2)-(5/9)*(I*3^(1/2)+1)*c/(-1+I*3^(1/2))+(50/27)*c^2+(38000/2187)*c^3/(-1+I*3^(1/2))-(1868750/19683)*c^4/(I*3^(1/2)+1)], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4]])

(2)

``


 

Download sol.mw

 

solve(15-(1/100)*m = 5+(1/600)*m, {m})
#{m = 6000/7}

solve(15-1/(100*m) = 5+1/(600*m), {m})
#{m = 7/6000}

Regards,Mariusz

Please read in Maple Help (Ctrl+F1 and put "How Do I,Solve an Ordinary Differential Equation?" in Search->Enter):

Scroll down to:

      -Solving an ODE Numerically

      - Taking Derivatives and Integrals of Numeric Solutions

      -  Why You Should Not Use Int or Diff on a Numeric Solution 

 

Corrected file:question-vers_2.mw

(Edited: 2 times) :P

    
 


 

restart

with(Student[Calculus1])

solve([-(1.25*y-sqrt(abs(x)))*abs(1, x)/sqrt(abs(x))+2*x, 3.1250*y-2.50*sqrt(abs(x))], [x, y])

[]

(1)

solX := solve(((125*(1/100))*y-sqrt(abs(x)))^2+x^2-1 = 0, {y})

{y = (4/5)*abs(x)^(1/2)+(4/5)*(-x^2+1)^(1/2)}, {y = (4/5)*abs(x)^(1/2)-(4/5)*(-x^2+1)^(1/2)}

(2)

c1 := CriticalPoints(rhs(solX[1, 1]), x)

[-1, -(1/12)*((215+12*321^(1/2))^(2/3)-(215+12*321^(1/2))^(1/3)+1)/(215+12*321^(1/2))^(1/3), 0, (1/12)*((215+12*321^(1/2))^(2/3)-(215+12*321^(1/2))^(1/3)+1)/(215+12*321^(1/2))^(1/3), 1]

(3)

evalf(c1)

[-1., -.5566930951, 0., .5566930951, 1.]

(4)

point1 := evalf(subs(x = 1, rhs(solX[1, 1])))

.8000000000

(5)

point2 := evalf(subs(x = 0, rhs(solX[2, 1])))

-.8000000000

(6)

point3 := evalf(subs(x = -.5566930951, rhs(solX[1, 1])))

1.261469543

(7)

c2 := CriticalPoints(rhs(solX[2, 1]), x)

[-1, 0, 1]

(8)

with(plots); p1 := pointplot([[evalf(c1)[2], point3], [evalf(c1)[4], point3], [0, point2], [0, point1]], color = [green], symbol = circle, axes = none, symbolsize = 15); p2 := implicitplot(((125*(1/100))*y-sqrt(abs(x)))^2+x^2-1, x = -1 .. 1, y = -1 .. 1.3, axes = normal, gridrefine = 3); display({p1, p2})

 

``


 

Download Critical_Points.mw

sol := n > ceil(evalf(solve(product(exp(1/i), i = 1 .. n) = 100, n))); solve(sol, n);

# sol := 56 < n

# RealRange(Open(56), infinity)

First 14 15 16 17 18 19 20 Page 16 of 20